Question 13.7: Determine the range of sampling interval, T, that will make ...

Since H(s) = 1, the z-transform of the closed-loop system, T(z), is found from Figure 13.10 to be

T (z) = \frac{G (z)}{1  +  G (z)}          (13.53)

To find G(z), first find the partial-fraction expansion of G(s).

G (s) = 10  \frac{1  –  e^{-Ts} }{s  (s  +  1)} = 10  (1  −  e^{-Ts}) \left(\frac{1}{s}   –   \frac{1}{s  +  1} \right)               (13.54)

Taking the z-transform, we obtain

G (z) = \frac{10  (z  −  1) }{z} \left[\frac{z}{z  –  1}   –   \frac{z }{z  –  e^{-T}} \right] = 10 \frac{(1  −  e^{-T} )}{(z  − e^{-T})}        (13.55)

Substituting Eq. (13.55) into (13.53) yields

T (z) = \frac{10  (1  −  e^{-T}) }{z  –  (11e^{-T}  −  10)}       (13.56)

The pole of Eq. (13.56), (11e^{-T}  − 10), monotonically decreases from +1 to −1 for 0 < T < 0.2. For 0.2 < T < ∞, (11e^{-T} − 10) monotonically decreases from −1 to −10. Thus, the pole of T(z) will be inside the unit circle, and the system will be stable if 0 < T < 0.2. In terms of frequency, where ƒ = 1/T, the system will be stable as long as the sampling frequency is 1/0.2 = 5 hertz or greater.